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(cos A + cos b)^2 + (sin A + sin B)^2=?

  • 2 years ago
  • 2 years ago

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  1. Yahoo! Group Title
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    @apoorvk @satellite73 @A.Avinash_Goutham @ajprincess @AndrewNJ

    • 2 years ago
  2. apoorvk Group Title
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    okay, expand the two whole-squares, it's simple enough...

    • 2 years ago
  3. Yahoo! Group Title
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    the options are 4cos^2(a-b/2) 4 sin a/2 cos b/2

    • 2 years ago
  4. Yahoo! Group Title
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    |dw:1339426035616:dw|

    • 2 years ago
  5. Yahoo! Group Title
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    @apoorvk after expanding i got this then next..............

    • 2 years ago
  6. apoorvk Group Title
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    great, now.. cosAcosB + sinAsinB = cos(A-B) Use this^

    • 2 years ago
  7. patmyer1 Group Title
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    ugh, trig identities are a drag. Even after deriving them myself I simply CANNOT seem to memorize them

    • 2 years ago
  8. AndrewNJ Group Title
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    spread it so we have (cosa)^2+(cosb)^2+2(cosa)(cosb)+(sina)^2+(sinb)^2+2(sina)(sinb) =2+2cos(a-b)

    • 2 years ago
  9. AndrewNJ Group Title
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    (cosa)^2+(sina)^2=1 (cosb)^2+(sinb)^2=1 (cosa)(cosb)+(sina)(sinb)=cos(a-b)

    • 2 years ago
  10. Yahoo! Group Title
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    ok @apoorvk so it is not over yet!!@

    • 2 years ago
  11. shivam_bhalla Group Title
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    Continuing from where @AndrewNJ left, \[2(1+\cos(a-b))\] now we know that \[1 + \cos 2a = 2 \cos^2 a\] Now try :D

    • 2 years ago
  12. Yahoo! Group Title
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    @shivam_bhalla did nt understand!

    • 2 years ago
  13. shivam_bhalla Group Title
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    = \[2(1+\cos 2\frac{(a-b)}{2})\] Now we have a identity, \[1 + \cos 2x = 2 \cos^2 x\] in your case, \[x = \frac{a-b}{2}\] Now try

    • 2 years ago
  14. Yahoo! Group Title
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    i got it thxx @shivam_bhalla

    • 2 years ago
  15. shivam_bhalla Group Title
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    Your welcome :)

    • 2 years ago
  16. Yahoo! Group Title
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    @shivam_bhalla plzz help http://openstudy.com/study?signup#/updates/4fd60c0be4b04bec7f170b7d

    • 2 years ago
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